Find the domain of:
$$f(x) = \frac{1}{(g+1)x^2 + 2(g-1)x + g-3}$$
for the various values of $g\in \mathbb{R}$
I am trying to solve this. First of all, I take the $g$ value equal to $-1$ so the domain is $\mathbb{R}$.
The next thing that I need to do is to take the g value not equal to $-1$.
The discriminant must be greater than zero. After many calculations, the result I have is this:
$$-3g^2+6g+13>0$$
but I think that is not correct. This is an exercise from a Greek book and it gives me the solution, but not step by step :
Domain of $f = R - \left[{\dfrac{3-g}{g+1}, -1 }\right]$
Any ideas?
Thanks!
The domain of the function is the set of $x$ values where the function is defined on. Looking at the expression, this is when the denominator is not zero. To find when this occurs, let's first do the complement: find when the denominator is zero. This means solving the following equation for $x$, $$(g+1)x^2 + 2(g-1)x + g-3 = 0$$
This is a quadratic equation, so you can use the standard formula, and fortunately the discriminant becomes a neat perfect square. After some simplification, you should get two solutions for $x$: $$\dfrac{3-g}{g+1},\; -1$$
These are the only two $x$ values for which the denominator of $f(x)$ becomes zero. Hence, the domain of the function is the real line excluding these two points.
As an aside, $\left[{\dfrac{3-g}{g+1}, -1 }\right]$ conventionally means the interval from $\dfrac{3-g}{g+1}$ to $-1$. Whereas $\left\lbrace \dfrac{3-g}{g+1}, -1 \right\rbrace$ conventionally means the set of the two points $\dfrac{3-g}{g+1}$ and $-1$, which is what is needed for the solution.